# Join prime pseudovarieties

A pseudovariety $$\mathbf{V}$$ of groups is join prime if for any pseudovarieties $$\mathbf{V}_1, \mathbf{V}_2, \ldots,\mathbf{V}_m$$, the implication $$\mathbf{V} \subseteq \mathbf{V}_1 \vee \mathbf{V}_2 \vee \cdots \vee \mathbf{V}_m \quad \Longrightarrow \quad \mathbf{V} \subseteq \mathbf{V}_i$$ holds for some $$i$$. A finite group is join prime if it generates a join prime pseudovariety.

It is known that all groups of order up to 7 are join prime. So it is natural to ask: is the dihedral group $$D_4$$ of order 8 join prime?

Yes, the pseudovariety generated by $$D_4$$ is join prime (and the argument shows that the same is true for the pseudovariety generated by $$8$$-element quaternion group). The result follows from two observations:

(1) the class $${\mathbf P}$$ of finite groups whose Sylow $$2$$-subgroups are abelian forms a pseudovariety (i.e., this class is closed under finite products, the formation of subgroups and the formation of quotients), and
(2) any pseudovariety not contained in $${\mathbf P}$$ contains $$D_4$$ (and $$Q_8$$).

Assuming Items (1) and (2), and the obvious fact that $$D_4\not\in {\mathbf P}$$, we argue as follows: if $${\mathbf V}(D_4)\subseteq {\mathbf V}_1\vee \cdots \vee {\mathbf V}_m$$, then by Item (1) there is some $$i$$ such that $${\mathbf V}_i\not\subseteq {\mathbf P}$$. By Item (2), $${\mathbf V}_i$$ contains $$D_4$$, so $${\mathbf V}(D_4)\subseteq {\mathbf V}_i$$.

I explain how to prove Item (2). Assume $${\mathbf V}$$ is a pseudovariety containing some group $$G$$ with a nonabelian Sylow $$2$$-subgroup. We may assume that $$G$$ is chosen with $$|G|$$ minimal, and that $${\mathbf V}={\mathbf V}(G)$$. Necessarily $$G$$ is a nonabelian, subdirectly irreducible $$2$$-group with monolith $$M = \langle z\rangle\subseteq Z(G)$$ where $$z^2=1$$, and $$G/M$$ is abelian. In particular, $$G$$ is $$2$$-step nilpotent.

Since $$G/M\models [x,y]\approx 1$$ and $$M\models x^2\approx 1$$ we get that $$G\models [x,y]^2\approx 1$$.

It follows from commutator collection that the set of laws of any finite $$2$$-step nilpotent group may be axiomatized by: the group laws, the law $$[[x,y],z]\approx 1$$, an exponent bound $$x^m\approx 1$$, and an exponent bound on the commutator subgroup $$[x,y]^n\approx 1$$. Thus, if the exponent of our group $$G$$ is $$2^r$$, then since $$G$$ is nonabelian and satisfies $$[x,y]^2\approx 1$$ we get that $${\mathbf V}(G)$$ is exactly the class of all finite, $$2$$-step nilpotent $$2$$-groups satisfying $$[x,y]^2\approx 1$$ and $$x^{2^r}\approx 1$$. If $$2^r\geq 4$$, this class contains $$D_4$$.

But we must have $$2^r\geq 4$$, since otherwise $$2^r\mid 2$$ and $$G\models x^2\approx 1$$. This can't happen since $$G$$ is nonabelian.